the area of a triangle is 1/2 *b * h

you know that the base is 7.

the height is between 3 and 4. (tricky...look at how the triangle is tilted if you draw a line in the middle of the triangle)

so the area has to be less than 14 and the only choice is 12.5

Here is the alternative approach for the previous one.


find lenghts of each sides of traingles.

QP = 5
PR = 5
QR = sqrt(50)

since (QR)^2 = (QP) ^ 2 + (PR) ^ 2
Its an right angle triangle

A = 1/2 * B * H = 1/2* QP * PR = 1/2* 5 * 5 = 25/2 = 12.5

wawatan wrote:the area of a triangle is 1/2 *b * h

you know that the base is 7.

the height is between 3 and 4. (tricky...look at how the triangle is tilted if you draw a line in the middle of the triangle)

so the area has to be less than 14 and the only choice is 12.5

why does the area have to be less than 14? Because the height's between 3 and 4?

find lenghts of each sides of traingles.

But how do you get that QR = sqrt(50) without knowing its a right triangle??

And if it is a right triangle why is the height PR (5)?? I just dont see that in the picture

distance1 = sqroot[(0 - 3)^2 + (4 - 0)^2]

distance2 = sqroot[(0 - 4)^2 + (4 - 7)^2]

distance 1 = sqroot25 (base)

distance 2 = sqroot25 (height)

5 x 5 / 2 = 12.5

You can prove that this is a right triangle.

If you look at the region underneath QP, you can see that it is a right triangle with hypotenuse 5 and base 4 and height 3. If you drop a line from R, then you will have another right triangle underneth PR, with hypotenuse 5 and base 3 and height 4. If you "close" the two triangles, they will form a box with base 4 and height 3. When you open them up, they form another right triangle, with base and height of 5 and hypotenuse of 5sqroot2 (this is also the value you would get if you used the distance formula). Notice also that the subject triangle is a special triangle in the form x : x : xsqroot2.